Question
In a survey of 3,236 adults, 46% said that they eat cereal for breakfast at least twice a week. Determine the margin of error for this study. Round it to the nearest hundredth. Interpret its meaning in context. (2 points) The margin of error is □ %. The survey reports with 95% confidence that between □°% and □ % of adults eat cereal for breakfast at least twice a week.
Round the margin or error and Theo ther numbers to the nearest whole
Round the margin or error and Theo ther numbers to the nearest whole
Answers
GPT-4o mini
To determine the margin of error for a proportion, we can use the formula:
\[
ME = Z \times \sqrt{\frac{p(1 - p)}{n}}
\]
Where:
- \( ME \) is the margin of error
- \( Z \) is the Z-score corresponding to the desired confidence level (for 95% confidence, \( Z \) is approximately 1.96)
- \( p \) is the sample proportion (in decimal form)
- \( n \) is the sample size
Given:
- \( n = 3,236 \)
- \( p = 0.46 \)
Let's plug in the values into the formula:
1. Calculate \( p(1 - p) \):
\[
p(1 - p) = 0.46 \times (1 - 0.46) = 0.46 \times 0.54 = 0.2484
\]
2. Calculate the standard error:
\[
SE = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.2484}{3236}} \approx \sqrt{0.0000768} \approx 0.00877
\]
3. Calculate the margin of error:
\[
ME = Z \times SE = 1.96 \times 0.00877 \approx 0.0172
\]
4. Convert the margin of error to a percentage:
\[
ME \approx 0.0172 \times 100 \approx 1.72\%
\]
Now, rounding to the nearest hundredth, the margin of error is approximately **1.72%**.
Next, we can calculate the confidence interval:
- The lower endpoint is:
\[
p - ME = 0.46 - 0.0172 = 0.4428 \approx 0.443 \text{ (rounded to the nearest whole percent)}
\]
- The upper endpoint is:
\[
p + ME = 0.46 + 0.0172 = 0.4772 \approx 0.477 \text{ (rounded to the nearest whole percent)}
\]
Thus, rounding those values back to percentage:
- Lower limit: Approximately **44%**
- Upper limit: Approximately **48%**
Putting it all together:
- The margin of error is **1.72%**.
- The survey reports with 95% confidence that between **44%** and **48%** of adults eat cereal for breakfast at least twice a week.
Interpretation: The results suggest that if similar surveys were conducted repeatedly, we would expect that between **44% and 48%** of adults in the population eat cereal for breakfast at least twice a week, with 95% confidence in this estimate.
\[
ME = Z \times \sqrt{\frac{p(1 - p)}{n}}
\]
Where:
- \( ME \) is the margin of error
- \( Z \) is the Z-score corresponding to the desired confidence level (for 95% confidence, \( Z \) is approximately 1.96)
- \( p \) is the sample proportion (in decimal form)
- \( n \) is the sample size
Given:
- \( n = 3,236 \)
- \( p = 0.46 \)
Let's plug in the values into the formula:
1. Calculate \( p(1 - p) \):
\[
p(1 - p) = 0.46 \times (1 - 0.46) = 0.46 \times 0.54 = 0.2484
\]
2. Calculate the standard error:
\[
SE = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.2484}{3236}} \approx \sqrt{0.0000768} \approx 0.00877
\]
3. Calculate the margin of error:
\[
ME = Z \times SE = 1.96 \times 0.00877 \approx 0.0172
\]
4. Convert the margin of error to a percentage:
\[
ME \approx 0.0172 \times 100 \approx 1.72\%
\]
Now, rounding to the nearest hundredth, the margin of error is approximately **1.72%**.
Next, we can calculate the confidence interval:
- The lower endpoint is:
\[
p - ME = 0.46 - 0.0172 = 0.4428 \approx 0.443 \text{ (rounded to the nearest whole percent)}
\]
- The upper endpoint is:
\[
p + ME = 0.46 + 0.0172 = 0.4772 \approx 0.477 \text{ (rounded to the nearest whole percent)}
\]
Thus, rounding those values back to percentage:
- Lower limit: Approximately **44%**
- Upper limit: Approximately **48%**
Putting it all together:
- The margin of error is **1.72%**.
- The survey reports with 95% confidence that between **44%** and **48%** of adults eat cereal for breakfast at least twice a week.
Interpretation: The results suggest that if similar surveys were conducted repeatedly, we would expect that between **44% and 48%** of adults in the population eat cereal for breakfast at least twice a week, with 95% confidence in this estimate.